Simple proof that $8\left(\frac{9}{10}\right)^8 > 1$

For this particular example, it's pretty easy, since $$ (9/10)^8 \geq (8/10)^4 \geq (6/10)^2 \geq (3/10) $$ and so you're done. Notice, my bounds are quite crude in the last two cases.

Repetitive squaring is the quickest way to get up to a large power. In fact, that's how a typical computer implementation will do it for integer powers.

Note that $(1-0.1)^8\geq 1-0.8$ by Bernoulli's inequality, as mentioned in a comment by lvb. For other cases you can use $\left(1-\frac{1}{n+1}\right)^k\geq 1-\frac{k}{n+1}$, which makes it easier to find a sufficient condition on $n$ for $k\left(\frac{n}{n+1}\right)^k$ to be larger than $1$.

In the motivating problem, the idea was that $k$ is fixed (at $8$), and since $\frac{n}{n+1}$ goes to $1$ as $n$ goes to infinity, $k\left(\frac{n}{n+1}\right)^k$ approaches $k$, and in particular is eventually bigger than $1$. One then finds out in a particular case what "eventually" means; here $n=9$ suffices. In general, Bernoulli's inequality leads to the conclusion that $n>k$ suffices.

You can use the fact that $(1 + x/n)^n$ approaches $e^x$ for large $n$. Then $$ \left(\frac{n-1}{n}\right)^{k} = \left(1 - \frac{1}{n}\right)^{k} \thicksim e^{-k/n} $$ as $n\rightarrow\infty$. In your case, this would give $(9/10)^8 \approx e^{-4/5} > 1/e$, which is clearly greater than $1/8$. (For $n=10$ and $x=-1$, the error in the exponential approximation is about 5%.)